A skater of mass 45.0 kg standing on ice throws a stone of mass 7.65 kg with a speed of 20.9 m/s in a horizontal direction. Find: a. The speed of the skater after throwing the stone.
b. The distance over which the skater will move in the opposite direction if the coefficient of kinetic friction between his skates and the ice is 0.03.

(a) 3.553 m/s

(b) 21.46 m

Explanation:

(a) Applying the law of of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u'  = mv+m'v' Equation 1

Where m and m' are the mass of skater and stone respectively,  u and u' are the initial velocity of skater and stone respectively, v and v' are the final velocity of the skater and the stone respectively.

Note, u = 0 m/s, u' = 0 m/s

Therefore,

0 = mv+m'v'

-mv = m'v' Equation 2

make v the subject of the equation

v = -m'v'/m Equation 3

Given: m = 45 kg, m' = 7.65 kg, v' = 20.9 m/s

Substitute into equation 3

v = 7.65(20.9)/45

v = -3.553 m/s

Hence the speed of the skater = 3.553 m/s

(b) F = mgμEquation 4

But F = ma

Therefore,

ma = mgμ

a = gμ Equation 5

Where a = acceleration of the skater, g = acceleration due to gravity, μ = coefficient of kinetic friction

Given: μ = 0.03, g = 9.8 m/s²

Substitute into equation 5

a = 0.03(9.8)

a = 0.294 m/s²

Using the equation of motion,

v² = u²+2as Equation 6

Where s = distance moved by the skater.

note that u = 0 m/s.

therefore,

v² = 2as

s = v²/2a Equation 7

Given: v = 3.553 m/s, a = 0.294

Substitute into equation 7

s = 3.553²/(2×0.294)

s = 12.62/0.588

s = 21.46 m

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